I want to find such function $p(n)$ in a way that obtaining one succes will be asymptotically-equivalent to obtaining two failures. I basically want to have something like this: $$ \lim_{n \to \infty}{\frac{{n \choose 1}p(n)(1-p(n))^{n-1}}{{n \choose 2}(1-p(n))^2p(n)^{n-2}}}= $$ $$= \lim_{n \to \infty}{\frac{np(n)(1-p(n))^{n-1}}{\frac{n!}{2(n-2)!}(1-p(n))^2p(n)^{n-2}}}=...=$$ $$= \lim_{n \to \infty} {\frac{2(1-p(n))^{n-3}}{(n-1)p(n)^{n-3}} } = 1$$ And now I'm kind of stuck... what do do with $n-1$ in denominator? I could use some hints.
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So you want to find function $p(n)$ s.t. getting $1$ success in $n$ trials with probability of success $p(n)$ is asymptotically the same as getting $2$ failures? – mihaild Nov 27 '20 at 21:00
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Yeah, that's exactly what I want. – Awerde Nov 27 '20 at 22:02
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We don't even need asymptotic here. Just write $$\frac{2(1 - p(n))^{n - 3}}{(n - 1)p(n)^{n - 3}} = 1$$ and solve w.r.t. $p(n)$: $$\sqrt[n - 3]{2} (1 - p(n)) = \sqrt[n - 3]{n - 1} p(n)$$ $$p(n) = \frac{\sqrt[n - 3]{2}}{\sqrt[n - 3]{n - 1} - \sqrt[n - 3]{2}}$$ and if we replace $1$ in the first equation with $1 + o(1)$, we get general solution $$p(n) = \frac{\sqrt[n - 3]{2}}{\sqrt[n - 3]{n - 1} - \sqrt[n - 3]{2} + o(1)}$$
mihaild
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